$h(x)=\begin{cases} \text{cos}(x)&\text{for }x<\pi \\\\ \text{sin}(x)&\text{for }x\geq\pi \end{cases}$ Find $\lim_{x\to \pi^+}h(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D The limit doesn't exist.
Answer: Notice that we were asked to find the one-sided limit, $\lim_{x\to \pi^+}h(x)$. This is the limit where $x$ -values approach $\pi$ from the right. Let's find the limit as $x$ approaches $\pi$ from the right. We will use the fact that $h(x)=\text{sin}(x)$ for $x$ -values greater than $\pi$. $\begin{aligned} &\phantom{=}\lim_{x\to \pi^+}h(x) \\\\ &=\lim_{x\to \pi^+}\text{sin}(x) \\\\ &=\text{sin}(\pi)&\gray{\text{Direct substitution}} \\\\ &=0 \end{aligned}$ In conclusion, we found that $\lim_{x\to \pi^+}h(x)=0$.